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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42 GPA: 3.82
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If x and y are positive integers, when x^4-y^4 is divided by 4, what is the remainder?
1) x-y is divisible by 4
2) x^2+y^2 is divisible by 4
==> If you modify the original condition and the problem, you get x^4-y^4=(x-y)(x+y)(x^2+y^2). Then, from 1)=2), the remainder of what are both divided by 4 is 0, hence unique, and sufficient. The answer is D.
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The question below is also a 5051-level question, and an inequality problem that ignores what is squared.

Is 1/x > 1/x^2?

1) x>0
2) 1>1/x

==> If you modify the original condition and the problem, and square the inequality equation, what is squared is always a positive number. Hence, it does not change the sign of inequality. So, if you multiply both sides by x^2, you get x>1?. Since there is 1 variable, D is highly likely to be the answer. Since the range of the question should include the range of condition in order for the condition to be sufficient, in the case of 1), the range of the question does not include the range of the condition, hence not sufficient. In the case of 2), if you multiply both sides of the equation by x^2, you get x2>x, and from x(x-1)>0, then x<0 or 1<x. From this, the range of the question does not include the range of the condition, hence not sufficient. In the case of 1) & 2), from x>1, the range of the question includes the range of the condition, hence sufficient. The answer is C.

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Math Revolution GMAT Instructor V
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Also, you must be wary of CMT 4(A) question below.

(integer) If a and b are integers, is anodd number?
1) a+b is an even number
2) ab is an odd number

==>In the original condition, there are 2 variables, so C is the answer. Through 1) & 2), a=b=odd, so yes, hence sufficient. The answer is C. This problem, too, is an integer problem, which is a key question, so if you apply CMT 4(A), and tackle 2), a=b=odd, hence yes, and sufficient. B is the answer.

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If x and y are prime numbers, what is the number of the different factors of x4y3?

1) xy=15
2) x and y are different

==>In the original condition, x and y are prime numbers, so if x and y are different, the number of different factors of x4y3, is (4+1)(3+1)=20. However, in the case of 2), the condition is sufficient because of the word “different”. In the case of 1), too, (x,y)=(3,5),(5,3) then the number both two different factors is 20, hence sufficient. The answer is D. This is a common 5051-level problem related with CMT 4(B)
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On the coordinate planes system there are line L and line M, y=3x+2 and y=kx, respectively. For which of the following value of k did the line L intersect with the line M?

I. -3 II.-1 III.3
A.none B.I only C. III only D.I & II E.II & III

==>For the two straight lines to intersect with each other, the slope has to be different. Hence, from y=kx, it has to be k≠3, so I & II is the answer. The answer is D.

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If |m-n|=3, is m>n?
1) mn<0
2) The distance between m and 0 is 2

==>In the original condition, there are 2 variables(m,n) and 1 equation(|m-n|=3), so D is highly likely to be the answer. In the case of 1), (m,n)=(2,-1),(-2,1), so yes and no coexist, hence not sufficient. In the case of 2), you get |m-0|=|m|=2, m=±2, then (m,n)=(2,-1),(-2,1), hence yes/no, so not sufficient.
Through 1) & 2), you get (m,n)=(2,-1),(-2,1), so yes/no, hence not sufficient. The answer is E.

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If $0.2 is commission for sales of$1,000, what percent of sales amount is the commission?

A.2% B.0.2% C.0.02% D.0.002% E.0.0002%

==>0.2/1,000=2/10,000=0.0002=0.02%. That is because %=1/100.
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If m and n are integer, n=?
1) 2^m*5^n=8/25
2) m+n=1

==> In the original condition, there are 2 variables (m, n). Therefore, C is most likely to be the answer. By solving con 1) and con 2), m=3 and n=-2, and hence it is sufficient. The answer is C. However, since it is an integer question, if you apply CMT 4 (A) to con 1), from 2^m*5^n=2^3*5^-2, you get m=3 and n=-2, and hence it is sufficient. Therefore, the answer is B.

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If x is a prime number, which of the following cannot be an integer?

A. x/2 B. 5x/3 C. 3x/7 D. 15x/6 E. x/6

==> A. x=2, B. x=3, C. x=7, D. x=2. E. impossible

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U and T can produce 10,000 units in x hours when working together at their constant rates. If U can produce 10,000 units in 4x/3 hours alone at the constant rate, in how many hours can T produce 10,000 units alone at the constant rate, in terms of x?

A. 5x/2 B. 4x C. 2x D. x E. x/2

==> In case of work rate questions, if it is “together and alone”, you solve it reciprocally. In other words, if you assume the time it takes for T to produce 10,000 units alone as t hrs, you get t=4x from $$1/(4x/3)+1/t=1/x$$. Therefore, B is the answer.
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If x-1, x+6, x+7 are 3 side lengths of a right triangle, what is the value of x?

A. 4 B. 5 C. 6 D. 7 E. 8

==> $$(x-1)^2+(x+6)^2=(x+7)^2$$ to $$x^2-2x+1+x^2+12x+36=x^2+14x+49$$.
From$$x^2-4x-12=0$$to (x-6)(x+4)=0, x=6, since -4 does not work, you get x=6.
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If 5 different positive integers have 3 as its median, is the average (arithmetic mean) of them greater than 5?

1) The greatest integer of them is 16
2) The smallest integer of them is 1

==> If you modify the original condition and the question, the sum of 5 integers>585=25?, and so there are 5 variables and 1 equation. Therefore, E is most likely to be the answer. However, if the question is “greater than”, you need to find the least value. By solving con 1) and con 2),
The least value of the sum becomes 1+2+3+4+16=26>25 yes, hence it is sufficient. The answer is C. However, this question is a key question, so you need to apply CMT 4 (A).
For con 1), the least value of the sum=1+2+3+4+16=26>25, hence yes, it is sufficient.
For con 2), 1+2+3+4+5=15<25 is no, 1+2+3+10+30=46>25 is yes, hence it is not sufficient. Therefore, the answer is A.
This question, related to CMT 4 (A), is 5051-level question in current GMAT.

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If n is the product of 3 consecutive integers, which of the following must be true?

I. a multiple of 2 II. a multiple of 3 III. a multiple of 4

A. I only B. II only C. III only D. I and II E. II and III

==> The product of 3 consecutive integers always become the multiple of 6, because the product always contains 3 and 2. Thus, in this question, it always becomes the multiple of 6 that contain 3 and 2, I and II are the answer. Therefore, the answer is D. III does not work because it becomes 1*2*3*=6, hence it cannot be the multiple of 4.
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Which of the following is the approximation of $$(4^4+8^6)/(4^8+16^8)$$?

A. $$2^-^1^2$$
B. $$2^-^1^4$$
C. $$2^-^1^6$$
D. $$2^1^2$$
E. $$2^1^4$$

==> Because of the word “approximation”, you get $$(4^4+8^6)/(4^8+16^8)= 8^6/16^8$$. Also, from $$8^6/16^8 =(2^3)^6/(2^4)^8=2^1^8/2^3^2=2^1^8^-^3^2=2^-^1^4$$,
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There are 6 numbers x, y, 2, 5, 9, and 10. What is the range of the numbers?
1) The average (arithmetic mean) of x and y is 5
2) 2<x<y<10

==> In the original condition, there are 2 variables (x, y), and in order to match the number of variables to the number of equations, there must be 2 equations, and therefore C is most likely to be the answer. By solving con 1) and con 2), you get x+y=10, which becomes (x,y)=(3,7),(4,6)… Then, it is always range=10-2=8, so it is unique and sufficient. However, this statistics question is one of the key questions. So if you apply mistake type 4(A), for con 1), from x+y=10 and (x,y)=(3,7),(1,9), you always get range=8,9, and hence it is unique and sufficient. For con 2), from 2<x<y<10, you always get range=10-2=8, and hence it is unique and sufficient.

Therefore, the answer is B. This is a question related to mistake type 4(A).
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2) The integer question below is also a 5051-level question.

Is n an even number?
1) n(n+1)/2 is an even number
2) n(n+2) is an even number

==> In the original condition, there is 1 variable (n), and therefore D is most likely to be the answer.
However, for con 1), you get n=3 no, n=8 yes, and hence it is not sufficient, and for con 2), in order to get n(n+2)=even, you need to get n=even, and hence yes, it is sufficient. Therefore, the answer is B.
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If x and y are positive integers, $$x^y^-^2$$=?

1) $$x^2=1$$
2) $$y^2=4$$

==> In the original condition, there are 2 variables (x), and in order to match the number of variables to the number of equations, there must be 2 equations, and therefore C is most likely to be the answer. By solving con 1) & con 2), you get x=-1, 1 and y=-2, 2. Since x and y are positive integers, only x=1 and y=2 are possible, and con 1) becomes $$1^y^-^2=x^2^-^2=x^0=1$$.

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If m and n are positive integers, is 10n+m divisible by 3?
1) n=1
2) m=2

==> If you modify the original condition and the question, “$$Is 10^n+m=3t$$? (t=any positive integer)” is equal to “Is the sum of all the digits of $$10^n+m$$ divisible by 3?” Then, as from con 2), if you know m=2, from $$10^n+2$$ => 12, 102, 1002…., the sum of the digits always become 3, which is divisible by 3, and hence yes, it is sufficient.

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If x and y are positive integers, are they consecutive?
1) x+y=3
2) x-y=1

==> In the original condition, there are 2 variables (x, y), and in order to match the number of variables to the number of equations, there must be 2 equations as well. Therefore, C is most likely to be the answer. By solving con 1) and con 2),
For con 2), it is always yes, hence it is sufficient.
For con 1), only (x,y)=(1,2), (2,1) satisfies, hence it is yes.
This question is also related to CMT 4(B). The answer is D.

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If a and b are integers, is (ab+2)(ab+3)(ab+4) divisible by 12?

1) a=even
2) b=odd

==> If you modify the original condition and the question, since (ab+2)(ab+3)(ab+4) is the product of 3 consecutive integers, so it is always divisible by 3. Then, since (ab+2)(ab+3)(ab+4) is divisible by 3, in order for it to be divisible by 12, it needs to be divisible by 4, so you must get ab+2=even. For con 1), if a=even, you get ab=even, then you get ab+2=even and ab+4=even, so it is always divisible by 4, and hence it is yes. The answer is A. It is a CMT 4(A) question.
_________________ Re: Overview of GMAT Math Question Types and Patterns on the GMAT   [#permalink] 15 Jan 2017, 18:48

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